Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/sodium benzoate buffer of a desired pH of 4.00? The starting concentration of the benzoic acid is 3.00 M and the molecular weight of sodium benzoate is 144.10 g/mol.
Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/sodium benzoate buffer of a desired pH of 4.00? The starting concentration of the benzoic acid is 3.00 M and the molecular weight of sodium benzoate is 144.10 g/mol
See your post above or below. I've forgotten where I worked it.
To prepare the benzoic acid/sodium benzoate buffer with a desired pH of 4.00, we need to use the Henderson-Hasselbalch equation. This equation relates the pH of a buffer system to the pKa of the weak acid and the ratio of its conjugate base to weak acid.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
pH = desired pH
pKa = -log(Ka) (Ka is the acid dissociation constant of the weak acid)
[A-] = concentration of the conjugate base (in this case sodium benzoate)
[HA] = concentration of the weak acid (in this case benzoic acid)
First, we need to calculate the ratio of [A-]/[HA] using the Henderson-Hasselbalch equation.
pH = pKa + log ([A-]/[HA])
4.00 = -log(6.50 x 10^-5) + log ([A-]/[HA])
Rearranging the equation to solve for the ratio [A-]/[HA]:
log ([A-]/[HA]) = 4.00 + log(6.50 x 10^-5)
log ([A-]/[HA]) = 4.00 + (-log(6.50 x 10^-5))
log ([A-]/[HA]) = 4.00 - (-log(6.50 x 10^-5))
log ([A-]/[HA]) = 4.00 - (-log(6.50) - log(10^-5))
log ([A-]/[HA]) = 4.00 - (-0.8129 + 5)
log ([A-]/[HA]) = 4.00 - 4.1871
log ([A-]/[HA]) = -0.1871
Next, we need to determine the value of [A-]/[HA] using the logarithm table or a calculator. Taking the antilog gives:
[A-]/[HA] = antilog(-0.1871)
[A-]/[HA] = 0.668
Now, we are ready to calculate the concentrations of sodium benzoate ([A-]) and benzoic acid ([HA]).
Let the initial concentration of benzoic acid be [HA]0 = 3.00 M
Since [A-]/[HA] = 0.668, we can set up the equation:
[A-] = 0.668[HA] = 0.668(3.00 M) = 2.00 M
Now that we have the concentrations, to prepare 0.500 L of the buffer solution, we can calculate the amount of benzoic acid and sodium benzoate needed:
mass of benzoic acid = [HA]0 x volume of solution = (3.00 M) x (0.500 L) = 1.50 mol
mass of sodium benzoate = [A-] x volume of solution x molar mass
= (2.00 M) x (0.500 L) x (144.10 g/mol) = 144.10 g
Therefore, to prepare 0.500 L of the benzoic acid/sodium benzoate buffer with a desired pH of 4.00, you will need 1.50 moles of benzoic acid and 144.10 grams of sodium benzoate.